Simplify and expand the following expression: $ \dfrac{5}{z - 8}- \dfrac{3}{4z - 36}- \dfrac{3}{z^2 - 17z + 72} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{3}{4z - 36} = \dfrac{3}{4(z - 9)}$ We can factor the quadratic in the third term: $ \dfrac{3}{z^2 - 17z + 72} = \dfrac{3}{(z - 8)(z - 9)}$ Now we have: $ \dfrac{5}{z - 8}- \dfrac{3}{4(z - 9)}- \dfrac{3}{(z - 8)(z - 9)} $ The least common multiple of the denominators is: $ (z - 8)(z - 9)$ In order to get the first term over $(z - 8)(z - 9)$ , multiply by $\dfrac{4(z - 9)}{4(z - 9)}$ $ \dfrac{5}{z - 8} \times \dfrac{4(z - 9)}{4(z - 9)} = \dfrac{20(z - 9)}{(z - 8)(z - 9)} $ In order to get the second term over $(z - 8)(z - 9)$ , multiply by $\dfrac{z - 8}{z - 8}$ $ \dfrac{3}{4(z - 9)} \times \dfrac{z - 8}{z - 8} = \dfrac{3(z - 8)}{(z - 8)(z - 9)} $ In order to get the third term over $(z - 8)(z - 9)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{3}{(z - 8)(z - 9)} \times \dfrac{4}{4} = \dfrac{12}{(z - 8)(z - 9)} $ Now we have: $ \dfrac{20(z - 9)}{(z - 8)(z - 9)} - \dfrac{3(z - 8)}{(z - 8)(z - 9)} - \dfrac{12}{(z - 8)(z - 9)} $ $ = \dfrac{ 20(z - 9) - 3(z - 8) - 12} {(z - 8)(z - 9)} $ Expand: $ = \dfrac{20z - 180 - 3z + 24 - 12}{4z^2 - 68z + 288} $ $ = \dfrac{17z - 168}{4z^2 - 68z + 288}$